检查输入是否为C中的整数类型-CheckifinputisintegertypeinC

12  

There are several problems with using scanf with the %d conversion specifier to do this:

使用%d转换说明符使用scanf有几个问题:

1. If the input string starts with a valid integer (such as "12abc"), then the "12" will be read from the input stream and converted and assigned to num, and scanf will return 1, so you'll indicate success when you (probably) shouldn't;

   如果输入字符串以一个有效整数(如“12abc”)开始，那么将从输入流中读取“12”，并将其转换为num，而scanf将返回1，因此当您(可能)不应该时，您将指示成功;

2. If the input string doesn't start with a digit, then scanf will not read any characters from the input stream, num will not be changed, and the return value will be 0;

   如果输入字符串不以数字开头，那么scanf将不会从输入流中读取任何字符，num将不会被更改，返回值将为0;

3. You don't specify if you need to handle non-decimal formats, but this won't work if you have to handle integer values in octal or hexadecimal formats (0x1a). The %i conversion specifier handles decimal, octal, and hexadecimal formats, but you still have the first two problems.

   您不指定是否需要处理非十进制格式，但如果必须处理八进制或十六进制格式(0x1a)的整数值，则不能这样做。%i转换说明符处理十进制、八进制和十六进制格式，但是仍然存在前两个问题。

First of all, you'll need to read the input as a string (preferably using fgets). If you aren't allowed to use atoi, you probably aren't allowed to use strtol either. So you'll need to examine each character in the string. The safe way to check for digit values is to use the isdigit library function (there are also the isodigit and isxdigit functions for checking octal and hexadecimal digits, respectively), such as

首先，您需要将输入读取为字符串(最好使用fgets)。如果你不允许使用atoi，你可能也不允许使用strtol。因此，您需要检查字符串中的每个字符。检查数字值的安全方法是使用isdigit库函数(还有分别检查八进制和十六进制数字的isodigit和isxdigit函数)，例如

while (*input && isdigit(*input))
   input++;    

(if you're not even allowed to use isdigit, isodigit, or isxdigit, then slap your teacher/professor for making the assignment harder than it really needs to be).

(如果你甚至都不允许使用isdigit、isodigit或isxdigit，那就扇你的老师/教授一巴掌，让你的作业比实际需要的更困难)。

If you need to be able to handle octal or hex formats, then it gets a little more complicated. The C convention is for octal formats to have a leading 0 digit and for hex formats to have a leading 0x. So, if the first non-whitespace character is a 0, you have to check the next character before you can know which non-decimal format to use.

如果您需要能够处理八进制或十六进制格式，那么它会变得有点复杂。C的约定是八进制格式的首位为0，十六进制格式的首位为0x。因此，如果第一个非空格字符是0，您必须先检查下一个字符，然后才能知道使用哪种非十进制格式。

The basic outline is

的基本轮廓

1. If the first non-whitespace character is not a '-', '+', '0', or non-zero decimal digit, then this is not a valid integer string;
2. 如果第一个非空白字符不是'-'、'+'、'0'或非零小数位数，那么这不是一个有效的整数字符串;
3. If the first non-whitespace character is '-', then this is a negative value, otherwise we assume a positive value;
4. 如果第一个非空白字符是'-'，则这是一个负值，否则我们假定为正值;
5. If the first character is '+', then this is a positive value;
6. 如果第一个字符是'+'，那么这是一个正值;
7. If the first non-whitespace and non-sign character is a non-zero decimal digit, then the input is in decimal format, and you will use isdigit to check the remaining characters;
8. 如果第一个非空格和非符号字符为非零小数位数，则输入为十进制格式，您将使用isdigit检查其余字符;
9. If the first non-whitespace and non-sign character is a '0', then the input is in either octal or hexadecimal format;
10. 如果第一个非空格和非符号字符为“0”，则输入为八进制或十六进制格式;
11. If the first non-whitespace and non-sign character was a '0' and the next character is a digit from '0' to '7', then the input is in octal format, and you will use isodigit to check the remaining characters;
12. 如果第一个非空格和非符号字符是“0”，而下一个字符是从“0”到“7”，则输入为八进制格式，您将使用isodigit检查其余字符;
13. If the first non-whitespace and non-sign character was a 0 and the second character is x or X, then the input is in hexadecimal format and you will use isxdigit to check the remaining characters;
14. 如果第一个非空格和非符号字符为0，第二个字符为x或x，则输入为十六进制格式，您将使用isxdigit检查其余字符;
15. If any of the remaining characters do not satisfy the check function specified above, then this is not a valid integer string.
16. 如果剩下的任何字符都不满足上面指定的check函数，那么这不是一个有效的整型字符串。

4  

First ask yourself how you would ever expect this code to NOT return an integer:

首先问问你自己，你怎么能指望这段代码不返回一个整数:

int num; 
scanf("%d",&num);

You specified the variable as type integer, then you scanf, but only for an integer (%d).

您将变量指定为类型integer，然后使用scanf，但只针对整数(%d)。

What else could it possibly contain at this point?

在这一点上它还可能包含什么?

0  

I looked over everyone's input above, which was very useful, and made a function which was appropriate for my own application. The function is really only evaluating that the user's input is not a "0", but it was good enough for my purpose. Hope this helps!

我查看了上面每个人的输入，这是非常有用的，并制作了一个适合我自己应用的函数。这个函数实际上只是评估用户的输入不是“0”，但它对我来说已经足够好了。希望这可以帮助!

#include

int iFunctErrorCheck(int iLowerBound, int iUpperBound){

int iUserInput=0;
while (iUserInput==0){
    scanf("%i", &iUserInput);
    if (iUserInput==0){
        printf("Please enter an integer (%i-%i).\n", iLowerBound, iUpperBound);
        getchar();
    }
    if ((iUserInput!=0) && (iUserInputiUpperBound)){
        printf("Please make a valid selection (%i-%i).\n", iLowerBound, iUpperBound);
        iUserInput=0;
    }
}
return iUserInput;
}

0  

Try this...

试试这个…

#include 

int main (void)
{
    float a;
    int q;

    printf("\nInsert number\t");
    scanf("%f",&a);

    q=(int)a;
    ++q;

    if((q - a) != 1)
        printf("\nThe number is not an integer\n\n");
    else
        printf("\nThe number is an integer\n\n");

    return 0;
}

0  

This is a more user-friendly one I guess :

我想这是一个更友好的方法:

#include

/* This program checks if the entered input is an integer
 * or provides an option for the user to re-enter.
 */

int getint()
{
  int x;
  char c;
  printf("\nEnter an integer (say -1 or 26 or so ): ");
  while( scanf("%d",&x) != 1 )
  {
    c=getchar();

    printf("You have entered ");
    putchar(c);
    printf(" in the input which is not an integer");

    while ( getchar() != '\n' )
     ; //wasting the buffer till the next new line

    printf("\nEnter an integer (say -1 or 26 or so ): ");

  }

return x;
}


int main(void)
{
  int x;
  x=getint();

  printf("Main Function =>\n");
  printf("Integer : %d\n",x);

 return 0;
}

0  

I developed this logic using gets and away from scanf hassle:

我使用get and away from scanf麻烦开发了这个逻辑:

void readValidateInput() {

    char str[10] = { '\0' };

    readStdin: fgets(str, 10, stdin);
    //printf("fgets is returning %s\n", str);

    int numerical = 1;
    int i = 0;

    for (i = 0; i <10; i++) {
        //printf("Digit at str[%d] is %c\n", i, str[i]);
        //printf("numerical = %d\n", numerical);
        if (isdigit(str[i]) == 0) {
            if (str[i] == '\n')break;
            numerical = 0;
            //printf("numerical changed= %d\n", numerical);
            break;
        }
    }
    if (!numerical) {
        printf("This is not a valid number of tasks, you need to enter at least 1 task\n");
        goto readStdin;
    }
    else if (str[i] == '\n') {
        str[i] = '\0';
        numOfTasks = atoi(str);
        //printf("Captured Number of tasks from stdin is %d\n", numOfTasks);
    }
}

0  

printf("type a number ");
int cOnverted= scanf("%d", &a);
printf("\n");

if( cOnverted== 0) 
{
    printf("enter integer");
    system("PAUSE \n");
    return 0;
}

scanf() returns the number of format specifiers that match, so will return zero if the text entered cannot be interpreted as a decimal integer

scanf()返回匹配的格式说明符的数量，如果输入的文本不能解释为十进制整数，则返回0

-1  

I was having the same problem, finally figured out what to do:

我遇到了同样的问题，终于想出了办法:

#include 
#include 

int main ()
{
    int x;
    float check;
    reprocess:
    printf ("enter a integer number:");
    scanf ("%f", &check);
    x=check;
    if (x==check)
    printf("\nYour number is %d", x);
    else 
    {
         printf("\nThis is not an integer number, please insert an integer!\n\n");
         goto reprocess;
    }
    _getch();
    return 0;
}

-1  

I've been searching for a simpler solution using only loops and if statements, and this is what I came up with. The program also works with negative integers and correctly rejects any mixed inputs that may contain both integers and other characters.

我一直在寻找一个更简单的解决方案，只使用循环和if语句，这就是我想到的。该程序还可以处理负整数，并正确地拒绝任何可能包含整数和其他字符的混合输入。

#include 
#include  // Used for atoi() function
#include  // Used for strlen() function

#define TRUE 1
#define FALSE 0

int main(void)
{
    char n[10]; // Limits characters to the equivalent of the 32 bits integers limit (10 digits)
    int intTest;
    printf("Give me an int: ");

    do
    {        
        scanf(" %s", n);

        intTest = TRUE; // Sets the default for the integer test variable to TRUE

        int i = 0, l = strlen(n);
        if (n[0] == '-') // Tests for the negative sign to correctly handle negative integer values
            i++;
        while (i  '9') // Tests the string characters for non-integer values
            {              
                intTest = FALSE; // Changes intTest variable from TRUE to FALSE and breaks the loop early
                break;
            }
            i++;
        }
        if (intTest == TRUE)
            printf("%i\n", atoi(n)); // Converts the string to an integer and prints the integer value
        else
            printf("Retry: "); // Prints "Retry:" if tested FALSE
    }
    while (intTest == FALSE); // Continues to ask the user to input a valid integer value
    return 0;
}

-1  

This method works for everything (integers and even doubles) except zero (it calls it invalid):

此方法适用于除零(它称其为无效)之外的所有事物(整数甚至双数):

The while loop is just for the repetitive user input. Basically it checks if the integer x/x = 1. If it does (as it would with a number), its an integer/double. If it doesn't, it obviously it isn't. Zero fails the test though.

while循环仅用于重复的用户输入。它检查整数x/x是否等于1。如果它这样做(就像它对待数字那样)，它就是一个整数/双精度数。如果没有，显然不是。Zero在测试中失败了。

#include  
#include 

void main () {
    double x;
    int notDouble;
    int true = 1;
    while(true) {
        printf("Input an integer: \n");
        scanf("%lf", &x);
        if (x/x != 1) {
            notDouble = 1;
            fflush(stdin);
        }
        if (notDouble != 1) {
            printf("Input is valid\n");
        }
        else {
            printf("Input is invalid\n");
        }
        notDouble = 0;
    }
}

-2  

I found a way to check whether the input given is an integer or not using atoi() function .

我找到了一种方法，可以使用atoi()函数检查给定的输入是否是整数。

Read the input as a string, and use atoi() function to convert the string in to an integer.

将输入读取为字符串，并使用atoi()函数将字符串转换为整数。

atoi() function returns the integer number if the input string contains integer, else it will return 0. You can check the return value of the atoi() function to know whether the input given is an integer or not.

函数的作用是:如果输入字符串包含整数，则返回整数，否则返回0。您可以检查atoi()函数的返回值，以知道给定的输入是否是整数。

There are lot more functions to convert a string into long, double etc., Check the standard library "stdlib.h" for more.

还有很多函数可以将字符串转换为长、双等，请查看标准库“stdlib”。h”。

Note : It works only for non-zero numbers.

注意:它只适用于非零数。

#include
#include

int main() {
    char *string;
    int number;

    printf("Enter a number :");
    scanf("%s", string);

    number = atoi(string);

    if(number != 0)
        printf("The number is %d\n", number);
    else
        printf("Not a number !!!\n");
    return 0;
}